A body of 20 g and +50 μC charge is suspended from the ceiling by a string. If a horizontal electric field of 10 kV/m is applied, the angle between the string and vertical direction is

(Take $g=10\, N/kg$)

$\tan^{-1} (2.50)$

The correct answer is Option (1) → $\tan^{-1} (2.50)$

Solution:

Given: mass $m=20\,g=0.02\,kg$, charge $q=50\,\mu C=50\times10^{-6}C$, electric field $E=10\,kV/m=10^4\,N/C$.

Electric force: $F=qE=50\times10^{-6}\times10^4=0.5\,N$

Weight: $W=mg=0.02\times9.8=0.196\,N$

At equilibrium: $\tan\theta=\frac{F}{W}=\frac{0.5}{0.196}\approx2.55$

$\theta=\tan^{-1}(2.55)\approx68.7^\circ$